at the points D and F make the angles FDE, DFE equal to two of the angles of the given triangle to which the inscribed one is to be similar, .. the angle at E will be equal to the third angle. Join AE, and produce it to G; and from G draw GH, GI respectively parallel to ED, EF; join HI. HIG is the triangle required. Since DE and EF are respectively parallel to HG, GI, the angle DEF is equal to HGI. Also DE HG :: AE: AG: EF: GI, whence (Eucl. vi. 6.) the triangles HG1, DEF are similar, and.. HGI is similar to the given triangle. (29.) In a given equilateral and equiangular pentagon, to inscribe a square. Let ABCDE be the given pentagon. Join EB; and from E draw EF perpendicular and equal to EB. Join AF; and from G, where it cuts ED, draw GH parallel to FE. Draw HI, GK parallel to EB. Join IK. HK is the square required. E H B Since HG is parallel to EF and HI to EB, K but EF=EB, .. HG=HI. And since AE=AB, :. (Eucl. vi. 2.) HE=IB; also GK and DC being parallel to EB, and DE=BC, .. EG= BK. The triangles EHG, IKB, therefore have two sides in each and the included angles equal, and .. HG=IK, and the angle EHG=BIK, whence HG and IK are also parallel; therefore also GK is equal to HI; hence the four sides are equal; and the angle at Hbeing a right angle, all the angles are right angles, and consequently HK is a square. (30.) In a given triangle to inscribe a rhombus, one of whose angles shall be in a given point in the side of the triangle. Let ABC be the given triangle, and D the given point. Join BD, and produce it; and with the centre A, and radius AC, describe a circle cutting it in E. Join AE; and draw DF parallel to it, FG parallel to AC, and GH to FD. FH is the rhombus required. B F G A C D 11 E Since FD is parallel to AE, BF : FD :: BA : AE; and since FG is parallel to AC, BF: FG :: BA : AC :: BA : AE, .. FD=FG; and the sides opposite to these are equal, .. the figure FDHG is a rhombus. (31.) To inscribe a circle in a given quadrant. B H Let ABC be the given quadrant. Bisect the angle ACB by the line CD; and at D draw DE touching the quadrant, and meeting CA produced in E. Make CF-AE. From F draw FG at right angles to AC. G is the centre of the circle required. C F From G draw GH perpendicular to BC. Join DF. Since the angle DCE is half a right angle, and the angle at D a right angle, DE=DC=AC=FE, . the angle EDF=EFD; whence also GDF=GFD, and GD=GF; and since the angles FCG, GCH are equal, and GC common to the right-angled triangles GFC, GHC, .. GF= GH; .. the three lines GD, GF, GH CC are equal and the circle described from the centre G, and distance of any one of them, will pass through the extremities of the other two, and touch the arc and sides in the points D, F, H, because the angles at those points are right angles. (32.) To describe a circle, the circumference of which shall pass through a given point, and touch a given straight line in a given point. Let AB be the given straight line, € the given point, in which the circle is to touch it, D the point through which it must pass. Draw CO perpendicular to B AB. Join CD; and at the point D make the angle CDO=DCO; the intersection of the lines CO and DO is the centre of the circle required. = Since the angle DCO=CDO, CO DO, and .. a circle described from the centre O, at the distance OD, will pass through C, and touch the line AB in C, bebecause OC is perpendicular to AB. (33.) To describe a circle which shall pass through given point, have a given radius, and touch a given straight line. Let AB be the given straight line, and C the given point through which the circle must pass. In AB take any point B; and from E A B it draw BD at right angles to AB, and equal to the given radius; through D draw DE parallel to AB; and with the centre C, and radius equal to the given radius, describe a circle cutting DE in O. O is the centre of the circle required. From O draw OF perpendicular to AB, it is equal to DB, i. e. to the given radius; and the circle described from the centre O, and radius OF, will touch (Eucl. iii. 16. Cor.) the line AB in F, and pass through C. (34.) To describe a circle which shall pass through two given points, and touch a given straight line. Let A, B be the given points, and CD the given straight line. Join AB. Join AB. And 1. let CD be parallel to AB. perpen C F A E B Join FA, and make the Bisect AB in E, and draw EF dicular to AB, and .. to CD. angle FAO=AFO; then will circle required. Since the angle FAO=AFO, A0=OF. But AE= EB, and EO is common to the triangles AEO, BEO, and the angles at E right angles, .. AO=OB. Whence AO, OB, OF are all equal; and the circle described from the centre O, at the distance of any one of them, will pass through the extremities of the other two, and touch the line CD, since OF is perpendicular to CD. 2. But if AB is not parallel to CD, let them be produced to meet in E; and take EF a mean proportional between - EA and EB. Join FA, FB; and describe a circle about the triangle AFB; it will be the circle required. E F D B Since EF is a mean proportional between EA and EB, EF touches the circle (Eucl. iii. 37.), which passes through A and B. (35.) To describe a circle, the circumference of which shall pass through a given point, and touch a circle in a given point; the two points not being in a tangent to the given circle. Chronicle 961 F E Let A be the given point in the circumference of the circle whose centre is 0; B the given point without. Join BA, and produce it to D. Join OD; and through A draw OAE; and draw BE parallel to OD, cutting OAE in E. E is the centre of the circle required. D Since (Eucl. i. 29.) the angle ODA is equal to ABE, and OAD to BAE, .. the triangles ODA, ABE are similar, and OD being equal to OA, AE will be equal to EB; a circle .. described with the centre E, and radius EA, will pass through B, and touch the circle ADF in the point A, since the line joining the centres passes through A. (36.) To describe a circle the centre of which may be in the perpendicular of a given right-angled triangle, and the circumference pass through the right angle and touch the hypothenuse. Let EAD be the given right-angled triangle, having the angle at A a right angle. Make EC EA. Join |